Problem with Parity for photons

[LayMuon]

I am trying to understand how it can be shown that under parity transformation we have to have \hat{P} \hat{a}_{\mathbf{p},\lambda} \hat{P} = - \hat{a}_{-\mathbf{p},\lambda}, I mean the negative sign (negative intrinsic parity of photon). So I am trying to prove that from the vector nature of four potential it follows that \eta = -1 in the relation \hat{P} \hat{a}_{\mathbf{p},\lambda} \hat{P} = \eta \hat{a}_{-\mathbf{p},\lambda}

In radiation gauge the second quantized four-potential is:
<br /> \mathbf{A}(x) = \int \frac{d^3p}{(2 \pi)^3 \sqrt{2 \omega_p}} \sum_{\lambda =1,2} \left [ \boldsymbol{\epsilon}(\mathbf{p}, \lambda) \hat{a}_{\mathbf{p},\lambda} e^{-ipx} +\boldsymbol{\epsilon}^*(\mathbf{p}, \lambda) \hat{a}_{\mathbf{p},\lambda}^* e^{ipx} \right ]<br />
From vector nature of $\mathbf{A}$ we should have:
<br /> \hat{P} \mathbf{A}(t,\mathbf{x}) \hat{P} = - \mathbf{A}(t,\mathbf{x&#039;}=-\mathbf{x})<br />

Applying parity operator:
<br /> \hat{P} \mathbf{A} \hat{P}= \int \frac{d^3p}{(2 \pi)^3 \sqrt{2 \omega_p}} \eta \sum_{\lambda =1,2} \left [ \boldsymbol{\epsilon}(\mathbf{p}, \lambda) \hat{a}_{-\mathbf{p},\lambda} e^{-ipx} +\boldsymbol{\epsilon}^*(\mathbf{p}, \lambda) \hat{a}_{-\mathbf{p},\lambda}^* e^{ipx} \right ] \\<br /> \hat{P} \mathbf{A}(x) \hat{P}= \int \frac{d^3p&#039;}{(2 \pi)^3 \sqrt{2 \omega_p}} \eta \sum_{\lambda =1,2} \left [ \boldsymbol{\epsilon}(-\mathbf{p&#039;}, \lambda) \hat{a}_{\mathbf{p&#039;},\lambda} e^{-ip&#039;x&#039;} +\boldsymbol{\epsilon}^*(\mathbf{-p&#039;}, \lambda) \hat{a}_{\mathbf{p&#039;},\lambda}^* e^{ip&#039;x&#039;} \right ]<br />

On the other hand for circular polarization vectors the flipping of the momentum sign ammouts to rotation of the direction of momentum by 180, e.g. around the axes $x$:
<br /> \boldsymbol{\epsilon}(\mathbf{p},1) = \frac{1}{\sqrt{2}}\{1,i,0\} \Rightarrow \boldsymbol{\epsilon}(-\mathbf{p},1) = \frac{1}{\sqrt{2}}\{1,-i,0\} \\<br /> \boldsymbol{\epsilon}(\mathbf{p},2) = \frac{1}{\sqrt{2}}\{1,-i,0\} \Rightarrow \boldsymbol{\epsilon}(-\mathbf{p},2) = \frac{1}{\sqrt{2}}\{1,i,0\}<br />

We see that polarizations exchange their places in the equation. Under spatial inversion all coordinates would be reversed and a minus sign would appear in the equation (4). Hence

<br /> \hat{P} \mathbf{A}(x) \hat{P}= \int \frac{d^3p&#039;}{(2 \pi)^3 \sqrt{2 \omega_p}} (-\eta) \sum_{\lambda =1,2} \left [ \boldsymbol{\epsilon}(\mathbf{p&#039;}, \mp) \hat{a}_{\mathbf{p&#039;},\pm} e^{-ip&#039;x&#039;} +\boldsymbol{\epsilon}^*(\mathbf{-p&#039;}, \mp) \hat{a}_{\mathbf{p&#039;},\pm}^* e^{ip&#039;x&#039;} \right ]<br />

I don't understand how can this equation (7) be compared with equation (2) to deduce that \eta=-1, the polarizations echanged their places!

 

[Bill_K]

Remember that, although momentum is a normal (polar) vector, spin is an axial vector. So that under a parity operation, p → -p but s → s. The components of the polarization vectors remain unchanged: ε(-p,1) is still (1/√2){1, i, 0}.

 

[LayMuon]

Remember that, although momentum is a normal (polar) vector, spin is an axial vector. So that under a parity operation, p → -p but s → s. The components of the polarization vectors remain unchanged: ε(-p,1) is still (1/√2){1, i, 0}.

That I intuitively realize but I need mathematically sound proof. Which part of my logic fails mathematically?